Calculus/calculus

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Question
These are related rate problems that I am having troule with. Hopefully you can help me.
A missile is launched following a trajectory of the function y=x^(1/3).
a) how fast is the missile moving along the ground when x=8?
b)how fast is the distance from the point of launch changing when x=8?
c)how fast is the angle of elevation changing when x=8?

Thanks!

Answer
(a) The speed at x=8 is y'(8).
Since y' = 1/[3x^(2/3)], at x=8 is is 1/[3*8^(2/3)].
Since the cube root of 8 is 2, the cube root squared is 4, so the bottom is 3*4 = 12.
That means the speed is 1/12.

(b) The distance is s(x,y) = sqrt(x+y) and y = x(1/3), so s(x) = sqrt(x^2 + x^(2/3)).
The change of this is dx/dx, which is (2x + 2[x^(-1/3)]/3)/sqrt(x^2 + x^(2/3)).
That is, dy/dx = 2[x + 1/x^(1/3)]/sqrt[x^2 + x^(2/3)].

Putting in x=9 gives 2(8 + 1/2)/sqrt(8 + 4).
That is 2(17/2)/sqrt(12).  The 2/2 on the top drops out, so we have 17/sqrt(12).
Multiplying top and bottom by sqrt(12) gives 17*sqrt(12)/12.

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